One of the points of intersection of the curves y = 1 + 3 x - 2 x 2 and y = 1 x is ( 1 2 , 2 ) . Let the area of the region enclosed by these curves be 1 24 ( l 5 + m ) - n log e ( 1 + 5 ) , where l , m , n ? N . Then l + m + n is equal to

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Solution

Q.
One of the points of intersection of the curves $y = 1 + 3 x - 2 x^{2}$ and $y = \frac{1}{x}$ is $(\frac{1}{2}, 2) .$ Let the area of the region enclosed by these curves be $\frac{1}{24} (l \sqrt{5} + m) - n \log_{e} (1 + \sqrt{5}),$ where $l, m, n \in N .$ Then $l + m + n$ is equal to [2024]

1 32

2 30

3 29

4 31

Ans.
(2)

Given curves are $y = 1 + 3 x - 2 x^{2} and y = \frac{1}{x}$

On solving, $2 x^{3} - 3 x^{2} - x + 1 = 0$

$\Rightarrow$ $(2 x - 1) (x^{2} - x - 1) = 0 \Rightarrow x = \frac{1}{2}, x = \frac{1 \pm \sqrt{5}}{2}$

Required Area = $\int_{\frac{1}{2}}^{\frac{1 + \sqrt{5}}{2}} (1 + 3 x - 2 x^{2} - \frac{1}{x}) d x$

$= {[x + \frac{3 x^{2}}{2} - \frac{2 x^{3}}{3} - \ln x]}_{\frac{1}{2}}^{\frac{1 + \sqrt{5}}{2}}$

$= [\frac{\sqrt{5} + 1}{2} + \frac{3}{8} {(\sqrt{5} + 1)}^{2} - \frac{1}{12} {(\sqrt{5} + 1)}^{3} - \ln (\frac{\sqrt{5} + 1}{2})] - (\frac{1}{2} + \frac{3}{8} - \frac{1}{12} - \ln \frac{1}{2})$

$= \frac{1}{24} [12 (\sqrt{5} + 1) + 9 {(\sqrt{5} + 1)}^{2} - 2 {(\sqrt{5} + 1)}^{3} - 12 - 9 + 2]$ $- (\ln \frac{\sqrt{5} + 1}{2} + \ln 2)$

$= \frac{1}{24} [12 (\sqrt{5} + 1) + 9 (6 + 2 \sqrt{5}) - 2 (5 \sqrt{5} + 1 + 3 \sqrt{5} (\sqrt{5} + 1) - 19)] - \ln (\frac{\sqrt{5} + 1}{2} \times 2)$

$= \frac{1}{24} [14 \sqrt{5} + 15] - \ln (\sqrt{5} + 1)$

$= \frac{1}{24} (l \sqrt{5} + m) - n \log_{e} (1 + \sqrt{5})$ [Given]

$∴$ $l = 14,$ $m = 15$ and $n = 1$

Hence, $l + m + n = 30$

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